Question 169638
Larger number={{{2x-1}}}
Smaller number={{{x}}}
Condition: The difference of the squares is 33:
{{{(2x-1)^2-x^2=33}}} -----------> EQN 1
{{{4x^2-4x+1-x^2=33}}}
{{{3x^2-4x+1-33=0}}}
{{{3x^2-4x-32=0}}},  solve via Quadratic Eqn: {{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}
where-------->{{{system(a=3,b=-4,c=-32)}}}
Continue and you get 2 values ---> {{{system(highlight(x=4),highlight(x=-8/3))}}}
Therefore, {{{x=4=smaller}}} & {{{2*4-1=8-1=7=Larger}}}, Answer
.
Check via EQn 1: 
{{{7^2-4^2=33}}}
{{{49-16=33}}}
{{{33=33}}}
Thank you,
Jojo